1. ## Logarithms

Suppose that for some values of A and r, P(t)=Aⅇ^(rt) and we know that P(3)=250 and P(11)=2P(0). Find P(7).

I have no clue how to even start this.

2. Originally Posted by velvet love
suppose that for some values of a and r, p(t)=aⅇ^(rt) and we know that p(3)=250 and p(11)=2p(0). Find p(7).

I have no clue how to even start this.
Given $\displaystyle p(11)=p(0)=a^0=1$

since $\displaystyle p(11) = a^{11r}=1$, $\displaystyle 11r=a\rightarrow r=\frac{a}{11}$

Plug $\displaystyle r=\frac{a}{11}$ into $\displaystyle P(t)$. You will get $\displaystyle P(t)=a^{\frac{a}{11}t}$

Now, $\displaystyle p(3)=a^{\frac{3a}{11}}=250$$\displaystyle \rightarrow\frac{3a}{11}= a^{250} Then take log of bothsides: \displaystyle \log{3a}-\log{11}= 250\log{a} You should be able to do the rest. You might want to check my algebra for I must go back to my work now. I'll catch up with you later. 3. I think i typed the original equation wrong. its supposed to be Ae^(rt). And p(11) = 2*p(0) 4. Originally Posted by Velvet Love I think i typed the original equation wrong. its supposed to be Ae^(rt). And p(11) = 2*p(0) Alright, let's try again. \displaystyle p(11)=2p(0) \rightarrow Ae^{11r}= Ae^0 \rightarrow$$\displaystyle Ae^{11r}= A\rightarrow $$\displaystyle e^{11r}= 1\rightarrow \displaystyle 11r\ln e= ln1\rightarrow$$\displaystyle 11r= ln1\rightarrow$$\displaystyle r= \frac{\ln1}{11}$

Now plug r into and t= 7 or whatever number to find the value of the function.

I don't have very good eye. I might have made mistakes when in a hurry going out the door. You might want to double check and triple chech my algebra. It would be best you take some time to review and practice your logarithm again. I understand you are underpressure.